For the diprotic weak acid H2A: Ka1= 3.7x10^-6 Ka2= 6.2x10^-9 What is the pH of
ID: 775480 • Letter: F
Question
For the diprotic weak acid H2A: Ka1= 3.7x10^-6 Ka2= 6.2x10^-9 What is the pH of a 0.0700M solution of H2A? What are the equilibrium concentrations in H2A and A^2- in this solution? For the diprotic weak acid H2A: Ka1= 3.7x10^-6 Ka2= 6.2x10^-9 What is the pH of a 0.0700M solution of H2A? What are the equilibrium concentrations in H2A and A^2- in this solution? Ka1= 3.7x10^-6 Ka2= 6.2x10^-9 What is the pH of a 0.0700M solution of H2A? What are the equilibrium concentrations in H2A and A^2- in this solution?Explanation / Answer
H2A -------> HA- + H+ -----------> A2- + H+
the second dissociation constant is very small compared to Ka1. So, we can consider only the first dissociation constant.
Ka1 = x * x/ (0.07-x)
3.7 * 10-6 = x2 / (0.07-x)
x is ver small so, 0.07-x ~0.07
x = 0.51 *10-3
[H+] = 0.00051
pH = -log [H+] = -log [0.00051]= 3.29
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Ka2 = x*x/ (0.00051-x)
6.2x10-9 = x2 / (0.00051-x)
x is ver small so, 0.00051-x ~ 0.00051
6.2x10-9 = x2 / 0.00051
x = 1.78 * 10-6
[A2-] = 1.78 * 10-6 M
equilibrium concentration of H2A =0.07- 0.00051 = 0.069 M and A2- = 1.78 * 10-6 M
H2A HA- H+ initial 0.07 0 0 change -x +x +x equilibrium 0.07-x x x